In humans, differences in the ability to taste phenylthiourea are due to two autosomal alleles at a single locus. Inability to taste is recessive to ability to taste. A child who is a nontaster is born to a couple who can both taste the substance. What is the probability that their next child will be a taster? 3/8
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- As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyThe following pedigree shows the pattern of inheritance of red-green color blindness in a family. Females are shown as circles and males as squares; the squares or circles of individuals affected by the trait are filled in black. What is the chance that a son of the third-generation female indicated by the arrow will be color blind if the father is not color blind? If he is color blind?In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the probability that the first child born to the couple will either be a freckled, colorblind boy or a non‑freckled, normal visioned girl or a non-freckled, normal visioned boy? What is the probability that the first four children born to the couple will be freckled and normal visioned girls?
- In humans, the genetic disease cystic fibrosis is caused by a recessive allele (a). The normal (healthy) allele is dominant (A). What is the genotype of someone who has cystic fibrosis? What are the two different genotypes that a healthy person could have? If two people were both heterozygous for the cystic fibrosis gene, what fraction of their children would be likely to have this disease? Hint: Draw a Punnett square to figure it out.In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the genotype of the woman's father? What is the probability that the couple's first child will be a non-freckled, normal visioned girl? What is the probability that the first two children born to the couple will be freckled and colorblind girls?The ability to detect the bitter taste of Phenylthiocarbamide (PTC) is inherited. “Tasting” is controlled by a dominant allele, while “non-tasting” is recessive PTC paper placed in your mouth can determine if you are a “taster” or a “non-taster”? Let's assume that you are a "taster". What possible genotypes could produce your phenotype? If you produced children with a "non-taster", what genotypic and phenotypic ratios would be expected in your offspring?
- In cats, a sex-linked recessive allele, d, results in dystrophin-deficient muscular dystrophy. Its dominant counterpart, D, results in normal levels of dystrophin in muscle. Also, the dominant allele F at an autosomal locus is responsible for polydactyly (the presence of extra digits on each foot) and the expression of its recessive counterpart f results in the absence of polydactyly (having the 5 digits per front paw, and 4 digits per hind paw). A healthy female cat that has polydactyly gives birth to a male kitten with the normal number or toes. In addition, this male kitten has muscular dystrophy. What can you conclude about the genotype of: Unknown alleles should be indicated with a “?" The son? The father? The mother?Above is a pedigree for colorblindness. Based on the pedigree, is the disease dominant or recessive and is it sex-linked or autosomal? Why? Furthermore, what is the probability that 18 on this chart is affected but the condition, and what is the probability that 18 is a carrier? Why? Are the probability of being a carrier and an affected individual different? WhyIn rabbits, the color of body fat is controlled by a single genewith two alleles, designated Y and y. The outcome of this traitis affected by the diet of the rabbit. When raised on a standardvegetarian diet, the dominant Y allele confers white body fat,and the y allele confers yellow body fat. However, when raisedon a xanthophyll-free diet, a homozygote yy rabbit has whitebody fat. If a heterozygous rabbit is crossed to a rabbit withyellow body fat, what are the proportions of offspring withwhite and yellow body fat when raised on a standard vegetariandiet? How do the proportions change if the offspring are raisedon a xanthophyll-free diet?
- In general terms, genes found on the same chromosome are linked, and will appear to defy Mendel’s Law of Independent Assortment. This law states that alleles (Links to an external site.) for different traits (Links to an external site.) are transmitted (Links to an external site.) to offspring (Links to an external site.) independently of one another. Functionally, this means that in a dihybrid testcross, in which a heterozygote is crossed to a double homozygous recessive individual, the expected 1:1:1:1 ratio will not be obtained. Instead, lower than expected numbers of non-parentals will result, because these non-parental flies are the result of recombination during synapsis. Interestingly, and functionally important in this exercise, synapsis only occurs in female fruit flies, requiring that the heterozygote in any study of linkage must be the female. Determining the relative positions of linked genes on a chromosome can be accomplished by calculating the frequency of…In earlobes, free (F) earlobe is dominant while attached (f) earlobe is recessive. What proportion of the offspirng would we expect to have attached earlobes in the following crosses: 1. Heterozygous for free earlobes with heterozygous dominant for free earlobes. (Answer in fraction) ______ 2. Heterozygous dominant for ree earlobes with homozygous dominant for free earlobes. _____ 3. Homozygous dominant for free earlobes with attacjhd earlobes. ______Pea plants have pods that are smooth (inflated) or constricted in shape and green or yellow incolor. These two autosomal traits observe the laws of Mendelian inheritance and assortindependently of one another. One gene affects pod shape and one gene affects pod color. Theallele for the smooth shape of pods (I) is dominant over the allele for the constricted shape ofpods (i) and the allele for the green color of pods (G) is dominant over the allele for the yellowcolor of pods (g).A pure breeding female parental pea plant with smooth green pods is cross-fertilized with a purebreeding parental male pea plant with constricted yellow pods. The resulting F1 generationplants are then self-fertilized and F2 generation progeny are obtained.The distribution of the appearance or phenotypes of the pods of the F2 generation plants are______.a- 100% smooth green podsb- 100% constricted yellow podsc- 50% smooth green pods, 50% constricted yellow pods, 0% smooth yellow pods, 0%constricted green podsd-…